3.122 \(\int \frac{\csc (e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac{b (5 a+2 b) \sec (e+f x)}{3 a^2 f (a+b)^2 \sqrt{a+b \sec ^2(e+f x)}}-\frac{b \sec (e+f x)}{3 a f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f (a+b)^{5/2}} \]

[Out]

-(ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]]/((a + b)^(5/2)*f)) - (b*Sec[e + f*x])/(3*a*(a
+ b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (b*(5*a + 2*b)*Sec[e + f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b*Sec[e + f*x
]^2])

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Rubi [A]  time = 0.141761, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4134, 414, 527, 12, 377, 207} \[ -\frac{b (5 a+2 b) \sec (e+f x)}{3 a^2 f (a+b)^2 \sqrt{a+b \sec ^2(e+f x)}}-\frac{b \sec (e+f x)}{3 a f (a+b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{f (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-(ArcTanh[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]]/((a + b)^(5/2)*f)) - (b*Sec[e + f*x])/(3*a*(a
+ b)*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (b*(5*a + 2*b)*Sec[e + f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b*Sec[e + f*x
]^2])

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\csc (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{b \sec (e+f x)}{3 a (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{3 a+2 b-2 b x^2}{\left (-1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{3 a (a+b) f}\\ &=-\frac{b \sec (e+f x)}{3 a (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{b (5 a+2 b) \sec (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{3 a^2 (a+b)^2 f}\\ &=-\frac{b \sec (e+f x)}{3 a (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{b (5 a+2 b) \sec (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{(a+b)^2 f}\\ &=-\frac{b \sec (e+f x)}{3 a (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{b (5 a+2 b) \sec (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-(-a-b) x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{(a+b)^2 f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)}}\right )}{(a+b)^{5/2} f}-\frac{b \sec (e+f x)}{3 a (a+b) f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac{b (5 a+2 b) \sec (e+f x)}{3 a^2 (a+b)^2 f \sqrt{a+b \sec ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 5.33787, size = 108, normalized size = 0.85 \[ \frac{\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},1-\frac{a \sin ^2(e+f x)}{a+b}\right )+(a+b) \left (3 a \sin ^2(e+f x)-2 (2 a+b)\right )\right )}{6 a^2 f (a+b) \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^5*(a^2*Hypergeometric2F1[-3/2, 1, -1/2, 1 - (a*Sin[e + f*x]^2)/(a
+ b)] + (a + b)*(-2*(2*a + b) + 3*a*Sin[e + f*x]^2)))/(6*a^2*(a + b)*f*(a + b*Sec[e + f*x]^2)^(5/2))

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Maple [B]  time = 0.68, size = 5056, normalized size = 39.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(f*x + e)/(b*sec(f*x + e)^2 + a)^(5/2), x)

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Fricas [B]  time = 0.987878, size = 1341, normalized size = 10.56 \begin{align*} \left [\frac{3 \,{\left (a^{4} \cos \left (f x + e\right )^{4} + 2 \, a^{3} b \cos \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt{a + b} \log \left (\frac{2 \,{\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{a + b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \,{\left (3 \,{\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \,{\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, \frac{3 \,{\left (a^{4} \cos \left (f x + e\right )^{4} + 2 \, a^{3} b \cos \left (f x + e\right )^{2} + a^{2} b^{2}\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{-a - b} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) -{\left (3 \,{\left (2 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (5 \, a^{2} b^{2} + 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \,{\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^4*cos(f*x + e)^4 + 2*a^3*b*cos(f*x + e)^2 + a^2*b^2)*sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a
+ b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)) - 2*(3*(2*a^3*
b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b
)/cos(f*x + e)^2))/((a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^4 + 2*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3
+ a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f), 1/3*(3*(a^4*cos(f*x + e)^4 + 2*a
^3*b*cos(f*x + e)^2 + a^2*b^2)*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*co
s(f*x + e)/(a + b)) - (3*(2*a^3*b + 3*a^2*b^2 + a*b^3)*cos(f*x + e)^3 + (5*a^2*b^2 + 7*a*b^3 + 2*b^4)*cos(f*x
+ e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^4 + 2
*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)/(b*sec(f*x + e)^2 + a)^(5/2), x)